Am I missing something? If the timeseries are identical how would calculating any statistical comparison beyond a simple check that they are in fact identical be meaningful?
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Joshua ChanceApr 15 '11 at 22:13

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I'm not commenting on the value of the question. I'm saying that it is incoherent. Two identical time series will never diverge and will never revert so it seams to me that asking whether they're cointegrated is more suited for philosophy than math.
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Joshua ChanceApr 17 '11 at 1:09

So now when we Dickey-Fuller test residuals in something like $$\Delta \hat u_t = \gamma_0 + \gamma_1 \hat u_{t-1} + \epsilon_t$$ nothing will be significant and we won't find any co-integration.

I am not precisely schooled in this theory, so I'm not sure if this means these series can't be referred to as "co-integrated" (clearly they have the same drift) or if this is just a trivial case where the test fails,

Two integrated series $X_t$ and $Y_t$ are cointegrated if their linear combination (some, not any) $\alpha X_t+\beta Y_t$ is stationary. If you have $P(X_t=Y_t)=1$ for all $t$, then $P(\alpha X_t+\beta Y_t=(\alpha+\beta) X_t)=1$. So according to definition of cointegration $(\alpha+\beta) X_t$ should be stationary, which is identical to $X_t$ being stationary. And here we get the contradiction, since $X_t$ is integrated, hence not stationary.

This was a basic explanation why you received your result. However a lot depends on how the actual statistic is computed. For other statistics or their software implementations you might get that two identical series are cointegrated, but that will not mean that they are. Two identical time series are the degenerate case which no-one checks against, and with degenerate cases you can always get unexpected results.

The first paragraph is wrong, in particular the definition of cointegration. Stationarity needs not be achieved for any linear combination, but only for some.
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RyogiApr 13 '12 at 20:36

@Ryogi, well I would have used word any if I wanted to mean any. If you do not prepend word any, then it is assumed that linear combination is specific. At least that is the convention in mathematical texts. But I can see that it might seem ambiguous. I'll change the wording to eliminate that.
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mpiktasOct 11 '12 at 7:17

You didn't use any quantifier in the original answer. That's wrong. Using some is acceptable. A mathematical text would state that there exist $\alpha$ and $\beta$ such that $\alpha X + \beta Y$ is stationary.
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RyogiOct 16 '12 at 4:03

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More importantly, since the cointegrating linear combination in this case requires $\alpha = -\beta$, the conclusion that $(\alpha + \beta) X$ must be stationary does not lead to any contradiction as $\alpha + \beta$ equals zero and $X$ could be anything.
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RyogiOct 16 '12 at 5:01

If $alpha=-\beta$ then $P((\alpha+\beta)X=0)=1$. I.e. we have a constant, which technically is a stationary process with variance zero. We can always exclude this case.
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mpiktasOct 16 '12 at 6:59

If they are both stationary then model $Y_t$ or $X_t$ in levels (and nothing is wrong).

If one of the two is $I(1)$ (non-stationary for one level), then take differences to ensure stationarity.

If they are both non-stationary, and hence $I(1)$, then test for co-integration:

if the residuals are $I(0)$, then we speak of the presence of cointegration. Estimate then an ECM model: $Y_t = \beta_0 + \beta_1 X_t + \eta_t$ obtaining $\hat{\beta_0}$ and $\hat{\beta_1}$ and using it in: $\Delta Y_t = \Delta X_t'\phi - \psi(Y_{t-1}-\hat{\beta_0} - \hat{\beta_1}X_t) + \varepsilon_t$. When $\varepsilon_t \sim N(0,1)$ then both $\psi$ and $\phi$ are asymptotically valid.

if the residuals are $I(1)$ then we speak of spurious regression. In that case you should model both variables by taking the first differences.